Problem 32 - Thomas Calculus

Question

Each of Exercises \(31-36\) gives a function \(f(x),\) a point \(c,\) and a positive number \(\epsilon .\) Find \(L=\lim _{x \rightarrow c} f(x) .\) Then find a number \(\delta>0\) such that for all \(x\) $$ 0<|x-c|<\delta \quad \Rightarrow \quad|f(x)-L|<\epsilon $$ $$ f(x)=-3 x-2, \quad c=-1, \quad \epsilon=0.03 $$

Step-by-Step Solution

Answer

The limit \(L\) is 1, and \(\delta = 0.01\).

Step 1: Find the Limit

First, we need to find the limit of the function as \(x\) approaches \(c\). The function given is \(f(x) = -3x - 2\), and we are asked to find \(\lim_{x \to -1} f(x)\).Calculate the limit:\[L = \lim_{x \to -1} (-3x - 2) = -3(-1) - 2 = 3 - 2 = 1\]Therefore, \(L = 1\).

Step 2: Establish the Epsilon Condition

We are given \(\epsilon = 0.03\). According to the epsilon-delta definition of a limit, for \(0 < |x + 1| < \delta\), we want \(|f(x) - L| < \epsilon\). This translates to:\[|-3x - 2 - 1| < 0.03\]

Step 3: Solve the Inequality for the Delta Condition

We begin with the inequality:\[|-3x - 3| < 0.03\]Factor and simplify:\[3|x - 1| < 0.03\]Divide by 3:\[|x - 1| < 0.01\]

Step 4: Relate Back to the Delta Condition

We need to express \(|x - 1| < 0.01\) in terms of \(|x + 1| < \delta\), given that we substituted \(x \to -1\) initially.Originally, the expression is \(|x + 1| < \delta\). Notice that \(|x + 1| = |x - (-1)|\).Thus, our \(\delta = 0.01\)So we find that \(\delta = 0.01\) satisfies the epsilon condition.

Key Concepts

Limit of a Function

When approaching any calculus problem involving limits, it's crucial to understand what a "limit of a function" really means. In our scenario, we have a function represented as \( f(x) = -3x - 2 \), and we want to evaluate its behavior as \( x \) approaches the number \( c = -1 \).
Think of a limit as the value that a function \( f(x) \) gets closer to when \( x \) moves towards a specific point, without necessarily reaching that point.
Here's how to find the limit:

• Substitute the number \( c \) into the function \( f(x) \). For our problem, this means substituting \( x = -1 \).
• Solve the equation: \( \lim_{x \to -1} (-3x - 2) = -3(-1) - 2 = 3 - 2 = 1 \).

Therefore, as \( x \) approaches \(-1\), \( f(x) \) approaches \( 1 \). This is our limit value \( L \).
Understanding this process is essential, as it forms the foundational concept for more complex calculus topics, including continuity and differentiability.

Epsilon-Delta Proof

An essential concept in calculus is the epsilon-delta definition of a limit, which provides a precise mathematical framework to argue about the behavior of functions.
The idea behind an epsilon-delta proof is to show that you can make the function \( f(x) \) as close as you want to the limit \( L \) (which we calculated as \( 1 \)) by choosing \( x \) values close enough to \( c \) (in this case, \(-1\)).
Let's break this down:

• "Epsilon" (\( \epsilon \)), is a small positive number that represents how close the function \( f(x) \) should be to the limit \( L \). Our exercise specifies \( \epsilon = 0.03 \).
• "Delta" (\( \delta \)), is another small positive number that indicates how close \( x \) should be to \( c \), ensuring \( f(x) \) falls within the epsilon distance from \( L \).

For our exercise:

• We want \( |f(x) - L| = |-3x - 2 - 1| < 0.03 \).
• Solving this inequality leads to establishing \( |x + 1| < 0.01 \) as our delta condition.

Hence, \( \delta = 0.01 \) ensures \( f(x) \) lies within the accepted epsilon neighborhood of \( L \).
This rigorous approach helps ensure that our understanding of limits is precise and trustworthy.

Calculus Exercises

Solving calculus exercises like these often involve combining algebraic manipulation with theoretical concepts such as limits.
These problems can be broken down into manageable steps:

• Start by identifying the function \( f(x) \), the point \( c \), and the epsilon \( \epsilon \) you need to work with.
• Calculate the limit \( L \) of the function as \( x \) approaches \( c \), ensuring you understand each step of substitution and simplification.
• Use the epsilon-delta definition to establish conditions for \( \delta \).This involves algebraic manipulation to solve inequalities ensuring \( |x - c| < \delta \) results in \( |f(x) - L| < \epsilon \).

By mastering these steps, you gain deeper insight into how limits work, enhancing your problem-solving skills in calculus.
Furthermore, practicing with varied problems will solidify your understanding and make tackling more complex problems much more intuitive. This foundational skill is pivotal for progressing in calculus as it paves the way for tackling derivatives and integrals.